In pp. 353-356 (pp. 397-400 of the pdf) of Claude Rabuel's Commentaires sur la geometrie de M. Descartes, he attempts to prove that a certain construction Descartes uses in his Geometrie for a certain kind of Oval works as intended.
The oval in question is a "Cartesian Oval", which can be defined similar to an ellipse. Fix two foci $F$ and $G$, and fix real numbers $n$, $c$. Then the oval in question is the locus of points $C$ such that $FC+nFG=c$. An ellipse is the special case where $n=1$, so a Cartesian Oval is a generalization of an ellipse.
The construction in question works by first fixing the foci $F$ and $G$, putting $A$ at the midpoint of $FG$, putting $L$ on $FG$ such that $GL:FL::n:1$, and putting $K$ at the midpoint of $AL$. A Cartesian Oval is to be constructed that passes through $A$ — this fixes the value of $c$. To construct this oval, pivot a ruler $FE$ around $F$, as shown. The length of the ruler doesn't matter. Tie a string so that it starts at $E$, goes around your finger at $C$ on $FE$, goes around $K$, goes around your finger at $C$ again, and finally ends at $G$. Descartes claims that if you pivot the ruler and keep your finger such that the string is taut at all times, it will trace the required oval. Also, the string length must be chosen so that your finger is at A when the ruler is horizontal.
Descartes doesn't prove this, but Rabuel attempts a proof. I will give an outline here. For all the algebraic details, see the link above. In the proof, he first defines $b=FA=AG$ (these are equal because $A$ is the midpoint of $FG$). Then he does the strange step of defining $z$, $d$, and $e$ by $CF=b+z$ and $CG=b-\frac{e}{d}z$. ($\frac{e}{d}$ is really a single variable, but at the time this was written all quantities had to be homogenous, so two were needed.) At this point in the proof, $\frac{e}{d}$ is not necessarily constant. However, if the proof worked, then we would expect it to be constant and equal to $\frac{1}{n}$. That way $FA+nFG$ would be constant, which is what we are trying to show.
Then he uses the constant length of the string to express $CK$ as a first-degree polynomial in $z$. He applies the Pythagorean theorem to get $CM^2=CK^2-MK^2=CF^2-MF^2=CG^2-MG^2$. He reduces the equations $CK^2-MK^2=CF^2-MF^2$ and $CK^2-MK^2=CG^2-MG^2$ to two quadratic equations in $z$. Let's write these quadratics as $z^2+\beta_1 z + \gamma_1 =0$ and $z^2+\beta_2 z + \gamma_2 =0$.
Now here is the really weird part. He assumes that these quadratics are perfect squares (!) and matches their coefficients with the coefficients of $z^2-2gz+gg=0$. For each quadratic equation, he solves for $g$ using the linear term to get $g=-\beta_1/2$ and $g=-\beta_2/2$. In effect, he is finding the average of the two roots. Then he equates these two values! So he is assuming that these two quadratics have the same average of their roots just because they share one root! If two monic quadratics share both a root and the average of their roots, then they are equal. So in effect, he is really assuming that these two quadratics are equal, just because they are monic and share a root! After this step, everything falls our beautifully. But how can we save this step? What was Rabuel intending with this step? Was he using some forgotten 17-18th Century algebraic technique?